Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#include <cctype>
#include <algorithm>
#include <cmath>
#include <map>
#include <string>
#include <vector>

using namespace std;
typedef long long ll;

int a[200005],d[200005],ans[200005],f[200005];
int main()
{
    int n,i;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        d[a[i]]=i;
    }
    for(i=n-1;i>=1;i--)
    {
        if(ans[i+1]==0)
        {
            if(d[i]>d[i+1])
            {
                ans[i]=1;
                f[i]=i+1;
            }
        }
        else
        {
            int aa=d[i],bb=d[i+1],cc=d[f[i+1]];
            if((aa<bb&&bb<cc)||(cc<aa&&aa<bb)||(bb<cc&&cc<aa))
            {
                ans[i]=ans[i+1];
                f[i]=f[i+1];
            }
            else
            {
                ans[i]=ans[i+1]+1;
                f[i]=i+1;
            }
        }
    }
    printf("%d\n",ans[1]);
    return 0;
}